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Chapter 7: Coordinate Geometry

1. If the point A (x , y) is equidistant from B (4 , 2) and C (-2 , 4). Find the relation between x and y.

|AB| = \(\sqrt{(x-5)^{2}+(y-1)^{2}}\)

= \(\sqrt{x^{2}+16-8x+y^{2}+4-4y}\)

|AC| = \(\sqrt{(x+2)^{2}+(y-5)^{2}}\)

= \(\sqrt{x^{2}+4+4x+y^{2}+16-8y}\)

Since, |AB| = |AC|

\(\Rightarrow\) \(x^{2}+y^{2}-8x-4y+20\) = \(x^{2}+y^{2}+4x-8y+20\)

\(\Rightarrow\)      8y – 4y = 4x + 8x

\(\Rightarrow\)      y = 3x

 2. Find the perimeter of a triangle with vertices (0, 8), (0, 0) and (6, 0)

Let the vertices of the \(\bigtriangleup\) be P(0, 8), Q(0, 0) and R(6, 0)

∴ PQ = \(\sqrt{(0)^{2}+(-8)^{2}}=\sqrt{64}= 8\)

∴ QR = \(\sqrt{(6)^{2}+(0)^{2}}=\sqrt{36}= 6\)

∴ RP = \(\sqrt{(-6)^{2}+(8)^{2}}=\sqrt{100}= 10\)

∴  Perimeter of \(\bigtriangleup\) = 8 + 6 + 10 = 24 units

3.Find the coordinates of the perpendicular bisector of the line segment joining the points P (1, 4) and Q (2, 3), cuts the y-axis.

Here, O (0, y), P (1, 4) and Q (2, 3)

AO = BO

\(\Rightarrow\) \(\sqrt{(0-1)^{2}+(y-4)^{2}}\) = \(\sqrt{(0-2)^{2}+(y-3)^{2}}\)

\(\Rightarrow\) \(1+y^{2}+16-8y= 4+y^{2}+9-6y\)

\(\Rightarrow\)   8y – 6y = 17 -13

\(\Rightarrow\)   2y = 4

                       y = 2

∴The required point is (0, 2)

4. The points which divides the line segment joining the points (5, 8) and (1, 2) in ratio 1:2 internally lies in which quadrant?

Now C\(\left ( \frac{1+10}{3}, \frac{2+16}{3} \right )\)

∴ C\(\left ( \frac{11}{3},6 \right )\)

Capture

Since, C\(\left ( \frac{11}{3},6 \right )\) lies in IV quadrant.

5. If A \(\left ( \frac{x}{2},3 \right )\) is the mid-point of the line segment joining the points B(4,1) and C(8,7). Find the value of x.

∴\(\frac{4+8}{2}=\frac{a}{2}\)

\(\Rightarrow\) 24 = 2a

\(\Rightarrow\) a = 12

 

6. A line intersects the y-axis at point A and B respectively. If (3, 4) is the mid-point of AB. Find the coordinates of A and B.

Here, \(\frac{x+0}{2}=3\) \(\Rightarrow x= 6\)

\(\frac{0+y}{2}=4\) \(\Rightarrow x= 8\)

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∴ The coordinates of A and B are (0, 8) and (6, 0).

7. Find the fourth vertex S of a parallelogram PQRS whose three vertices are P (-3, 8), Q (5, 4) and R (6, 2).

Here, \(\frac{-3+6}{2}=\frac{5+x}{2}\Rightarrow x=-1\)

\(\frac{8+2}{2}=\frac{4+x}{2}\Rightarrow x=3\)

∴The fourth vertices S of a parallelogram PQRS is S(-1, 3)

 8. Find the coordinates of the point which is equidistant from the three vertices of the

\(\bigtriangleup\)POQ.

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Point equidistant from the three vertices of  a right angle triangle is the mid-point of hypotenuse.

\(\Rightarrow \left ( \frac{4x+0}{2},\frac{0+4y}{2} \right )\Rightarrow (2x,2y)\)

 

9. Find the area of a triangle with vertices (x, y + z), (y, z + x) and (z, x + y).

Area of the required

\(\bigtriangleup = \frac{1}{2}|x(z+x-x-y)+y(x+y-y-z) +(y+z-z-x)|\)

= \(\frac{1}{2}|x(z-y)+y(x-z)+z(y-a)|\)

= \(\frac{1}{2}|0|\) = 0.

10. Find the area of a triangle with vertices P (2, 0), Q (6, 0) and R (5, 4)

Area of the required \(\bigtriangleup PQR\)

= \(\frac{1}{2}|2(0 – 2)+ 6(4 – 0)+ 5(0 – 0)|\)

= \(\frac{1}{2}|24-4|=10\) Sq. units.

11. If the points O (0, 0), P (3, 4), Q (x, y) are collinear, then write the relation between x and y.

Since, O (0, 0), P (3, 4), Q (x, y) are collinear.

\(\Rightarrow\) Area of a triangle formed by these points vanishes

∴ \( \frac{1}{2}|0(4 – y)+3(y-0)+x(0-4)|=0\)

\(\Rightarrow y-2x=0\Rightarrow 2x=y\)

 

12. Find the coordinates of the point O dividing the line segment joining the point P (6, 5)

and Q(2, 10) in the ratio 2:1.

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\(x=\frac{2(2)+1(6)}{5}= \frac{4+6}{5}= 2\)

and \(y=\frac{2(10)+1(5)}{5}= \frac{20+5}{5}= 5\)

∴ O(2, 5)

13. The line joining A(4, 3) and B(-3, 6) meets y-axis at C. At what ratio dose C divides the line segment AB?

Since,R lies on y-axis, let C be (0, y)

5

Suppose that the required ratio be z:1

∴\(x=\frac{-3z+4}{z+1}\Rightarrow 0= -3z+4\Rightarrow z=\frac{4}{3}\)

∴ The required ratio is 4 : 3.

  • Parmesh Gour

    find the value of y for which the distance between a point p[2,-3]andq[10,y]

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